A) 0
B) \[\pm \frac{3}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]
C) \[0,\,\pm \sqrt{\frac{3}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\]
D) \[0,\,\,\pm \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} a-x & c & b \\ c & b-x & a \\ b & a & c-x \\ \end{matrix}\, \right|\,=0\] \[\Rightarrow \] \[\left| \,\begin{matrix} a+b+c-x & c & b \\ a+b+c-x & b-x & a \\ a+b+c-x & a & c-x \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[(x-\sum{a})\,\left| \begin{matrix} 1 & c & b \\ 1 & b-x & a \\ 1 & a & c-x \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[x=\sum{a=0}\] (by hypothesis) or 1\[\{(b-x)\,(c-x)-{{a}^{2}}\}-c\{c-x-a\}+b\{a-b+x\}=0\] by expanding the determinant. or \[{{x}^{2}}-({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+(ab+bc+ca)=0\] or \[{{x}^{2}}-\left( \sum{{{a}^{2}}} \right)-\frac{1}{2}\,\left( \sum{{{a}^{2}}} \right)=0\] \[\left\{ \because \,a+b+c=0\Rightarrow {{(a+b+c)}^{2}}=0 \right.\] \[\Rightarrow \] \[\left. \sum{{{a}^{2}}}+2\sum{ab}=0\Rightarrow \sum{ab}=-\frac{1}{2}\sum{{{a}^{2}}} \right\}\] or \[x=\pm \sqrt{\frac{3}{2}\sum{{{a}^{2}}}}\] \[\therefore \] The solution is \[x=0\] or \[\pm \sqrt{\frac{3}{2}\sum{{{a}^{2}}}}\]. Trick: Put \[a=1,\,b=-1\] and \[c=0\] so that they satisfy the condition\[a+b+c=0\]. Now the determinant becomes \[\left| \,\begin{matrix} 1-x & 0 & -1 \\ 0 & -1-x & 1 \\ -1 & 1 & -x \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[(1-x)\,\{x(1+x)-1\}+1(1+x)=0\] \[\Rightarrow \] \[(1-x)\,\{{{x}^{2}}+x-1\}+x+1=0\] \[\Rightarrow \]\[x({{x}^{2}}-3)=0\] Now putting these in the options, we find that option (c) gives the same values i.e., 0, \[\pm \sqrt{3}\].
You need to login to perform this action.
You will be redirected in
3 sec
